Properties of Dedekind Domains
Throughout this article, A denotes a Dedekind domain.
Every fractional ideal of A can be written as
where each is a maximal ideal. The expression is unique up to permutation of terms.
In particular, the maximal ideals of A generate its Picard group.
Existence. First suppose is a non-zero ideal. If we are done. Otherwise it is contained in some maximal ideal , necessarily invertible. Now , because if equality holds multiplying both sides by gives , so . We replace by and repeat the process; this cannot continue indefinitely or we would have
Hence is a finite product of maximal ideals. Now a general fractional ideal is of the form for a non-zero ideal and . Write both and as a product of maximal ideals and we are done.
Uniqueness. Suppose , where (some possibly zero). If , after cancelling terms and reordering, we obtain a relation of the form:
This is an equality of ideals. We get a contradiction because the LHS is contained in but the RHS is not. ♦
Prove that an integral domain is a PID if and only if it is a Dedekind domain and a UFD. Hence we have the following Venn diagram.
Now we have the following.
Let M, N be non-zero fractional ideals of A. Write, as in proposition 1,
where each is maximal and . Then
Exercise. [ Localize at each ; reduce to the case where A is a dvr. ]
Constructing Dedekind Domains
How do we find Dedekind domains? For starters, a PID is a Dedekind domain so we have and (for any field k), but these are not too interesting.
Here is a standard way to construct new Dedekind domains from an existing one A. Let K be its field of fractions and L be a finite extension of K; we let B be the integral closure of A in L.
Prove that L is the field of fractions of B.
Now B has the following properties:
- it is a normal domain by construction;
- dim B = 1 since B is an integral extension of A, and dim A = 1 (by main theorem here).
Hence, if we can show that B is noetherian, then it is a Dedekind domain. But this result, called the Krull-Akizuki theorem, is surprisingly hard to prove. In most cases of interest, B will turn out to be a finitely generated A-module; thus B is noetherian as an A-module and hence as a module over itself.
1. Let be a finite extension and be the integral closure of in K. One can show that is a discrete subgroup of , when we topologize by identifying it with . Hence is a finitely generated -module; in particular it is a noetherian ring and hence a Dedekind domain. A more general result of this nature will be proven later.
2. Let k be any field such that . The integral closure of in is . [ Proof: exercise. ] Since A is clearly noetherian, it is a Dedekind domain.
Let be a square-free integer and , which is quadratic over . Prove that
Let . Each maximal gives a function
which takes to the exponent of in the above factorization of . This is called the valuation of at .
1. Prove the following properties of : for any we have:
Verify that if we set these properties hold for all .
2. Define a distance function on K by . Prove that (K, d) is a metric space, where the metric satisfies the strong triangular inequality.
A metric which satisfies this inequality is called an ultrametric. Interpret and prove the following statement: in an ultrametric space, every triangle is isoceles.
Valuation in Geometry
Suppose , a Dedekind domain. Recall that points P on the curve correspond bijectively to maximal ideals . Then is just the order of vanishing of f at P.
For example, let where ; for convenience write . Let us compute for :
Now since f does not vanish on P. From an earlier example, we have so and .
As a simple exercise, compute .
Recall in the last example here that for distinct maximal ideals of any ring A and we have a ring isomorphism
This gives the following.
If A is a Dedekind domain with finitely many maximal ideals , then it is a PID.
It suffices to prove that each is principal. Let us consider . Since we can pick . From the isomorphism
we can find such that
Then x is divisible by but not its square, and not divisible by . Thus . ♦
Let be a non-zero ideal; write
Then the composition factors of as an A-module comprise of copies of for . In particular
Recall that the composition factors of as an A-module are identical to those as a module over itself.
The above ring isomorphism
is in fact an isomorphism of A-algebras. Hence . We obtain a series of submodules of
with successive quotients . It suffices to show each of these is of dimension 1 over . To prove that, we recall that is a dvr so is a principal ideal. Since is already a module over we have:
as desired. ♦
Consider the special case where is the coordinate ring of a variety V over an algebraically closed field k. If A is a Dedekind domain, then since for each maximal ideal we have